Question

Match the LIST-I with LIST-II - Let $x_n = 3 + (-1)^n, n \in \mathbb{N}$. Then

Choose the correct answer from the options given below:

• A. A-I, B-IV, C-II, D-III
• B. A-I, B-III, C-II, D-IV
• C. A-IV, B-I, C-II, D-III
• D. A-IV, B-I, C-III, D-II

Hint:

An oscillating sequence with more than one limit point does not converge, so its limit does not exist!

Answer 1

The Correct Option is C

Step 1 : Concept:
This question involves analyzing sequence properties: supremum, infimum, boundedness, and existence of limits for oscillating sequences.

Step 2 : Key Formulas and Approach:
1. List terms of $x_n = 3 + (-1)^n$ for $n = 1, 2, 3, 4, \dots$
2. $\sup(x_n) = \text{Least Upper Bound}$.
3. $\inf(x_n) = \text{Greatest Lower Bound}$.

Step 3 : Step-by-step Explanation:

• Sequence Terms: For $n = 1$: $x_1 = 3 + (-1)^1 = 2$ For $n = 2$: $x_2 = 3 + (-1)^2 = 4$ For $n = 3$: $x_3 = 3 + (-1)^3 = 2$ For $n = 4$: $x_4 = 3 + (-1)^4 = 4$ The sequence set is $S = \{2, 4\}$.

• Item A:
$\sup x_n = \max\{2, 4\} = 4$. Matches with IV.

• Item B:
$\inf x_n = \min\{2, 4\} = 2$. Matches with I.

• Item C:
Since $2 \le x_n \le 4$ for all $n \in \mathbb{N}$, the sequence is bounded. Matches with II.

• Item D:
The sequence oscillates between $2$ and $4$, having two distinct limit points ($2$ and $4$). Hence, $\lim_{n \to \infty} x_n$ does not exist. Matches with III.

Step 4 : Final Answer:
The correct matching is A-IV, B-I, C-II, D-III, which corresponds to option (C).